text:
How is the planform geometry of a helicoptor rotor designed?
Rotor planform specially the tip is designed differently for different helicopter rotor. How the shapes are determined? For example I added 2 photos. How the dimensions of the tip or root of these rotors were determined?

First of all, a basic description of the aerodynamic environment encountered by the blades.
The airflow velocity $V$ seen by a rotating wing is the sum of:

the speed due to the rotation around the hub $=\omega R$;
the speed due to the forward velocity of the helicopter $=V_f$.

Those two speeds:

sum up when the blade in its rotation around the hub is rotating forward i.e. from the tailboom toward the nose and it's $V=\omega R + V_f$; at high forward speed, this sum can reach transonic value;
subtract  when the blade in its rotation around the hub is rotating backward i.e. from the nose toward the tailboom and it's $V=\omega R - V_f$; at high forward speed, this difference can reach very low subsonic speeds (it actually becomes negative on some 40% of the blade span toward the hub).

The following picture from this paper shows this effect (higher speeds on the right/advancing blade and lower speeds on the left/retracting blade):

That means that the velocity seen by blade continuosly changes from a maximum (the sum at point 1.) to a minimum (the difference at point 2.) during each revolution. In particular, the tip is the part of the blade experiencing the highest changes in speed and therefore the highest aerodynamic load.
There's a third aerodynamic aspect that is also important for  the design of the blade tip:

each blade leaves a wake behind it; this wake is stronger where the bladewise change of lift is stronger and this happens towards the tip; as seen (previous point 1.) toward the tips lift can be very high but then it suddenly goes to zero where the blade ends; this sudden change in lift leaves a very strong vortex behind, which is called tip vortex; now, when a blade bumps (literally) into the tip vortex released by the previous blade, the airflow is locally strongly perturbed (this is the source of the typical "flap-flap-flap" noise of slowly descending helicopters). The following picture from this paper shows two tip-vortices (number 3) shed by two blades:

So, now that we know the three main aerodynamic conditions the blade (and especially its tip) must undergo in its flight, we can try to figure it out how it should look like. Due to the previous point:

$\rightarrow$ the blade tip should be thin, with a short chord, with a leading edge swept back and a tapered planform, basically just like the wing of a jetliner; all this in order to diminish and/or retard the rise of drag due to transonic effects (wave drag);
$\rightarrow$ the blade tip should be thick, with a long chord and no particular need of being swept or tapered; actually if we want to be fancy to comply with this point we could use a delta planform for the tip: delta wing are known to generate quite a lot of lift at high AoA and low speeds, exactly the aerodynamic environment of 2.;
$\rightarrow$ the blade tip should be curved down (anhedral) so that the tip vortex is shed into the wake in a lower position than the following blade and therefore doesn't hit it directly.

Well, blade's design is driven by some quite conflicting requirements.
To all of this we must obviously add the centrifugal force stretching the blade: a blade of a typical EMS helicopter resists a centrifugal force of some 40kN (4000kg, 9000 pounds). To resist that force you need a structure which is as much as possible aligned to this centrifugal force i.e. you just want a perfectly straight and rectangular blade. Yet another conflicting requirement.
I let you figure out which requirement took over in the design of the blades in your picture.


text:
Which category is A321-271NX in?
I was wondering what the aircraft categories are all about because from my previous questions' answer I got to know about something called Cat D type aircraft. It made me curious to know what are the different types of aircraft categories and which one does the A321-271NX fall in, because it happens to be special for me!

The code is for the exact variant for this Airbus:

A321: This is an Airbus A321, the longest variant of the A320 family.

27*: It comes with PW1100 engines (which makes it a neo).

271: The exact engine variant, which can be found in the TCDS (Type Certificate Data Sheet):

PW1133G-JM Geared Turbo Fan jet engines (MOD 161002)

N: Added for all neo variants.

X: This means the aircraft has Cabin Flex, which allows the operator more freedom in choosing emergency exits:

these doors’ configuration flexibility are there to enable all customers’ operational needs without compromising on cabin comfort

(runwaygirlnetwork.com)

The "Cat D" you mentioned is the approach speed categorisation:

Category
Approach Speed

A
< 91 kt

B
91-120 kt

C
121-140 kt

D
141-165 kt

E
166-210 kt

The Airbus A321 can be in approach category C or D, depending on the exact variant:

A321[ceo] has Approach Category C to 75.5 tonnes MLW and D beyond that to 77.8 tonnes.

(SKYbrary - Airbus A321)
All A321neo aircraft are classified as Approach Category C:

ON A/C A321neo
[...]
The final approach speed is 136 kt at a MLW of 79 200 kg (174 606 lb) and classifies the
aircraft into the Aircraft Approach Category C.

(Airbus Aircraft Characteristics document AC A321 / A321neo)

Your last afirmation is not correct, only for A321Ceo no A321 NEO.
All A321neo are CAT C. See section "Final Approach Speed" - AIRCRAFT CHARACTERISTICS - AIRPORT AND MAINTENANCE PLANNING

Welcome to Aviation.SE! Thanks for the correction, I updated my answer ;) In the future, you can also suggest such corrections as an edit to the original answer. Have a look at the tour to see how the site works.

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text:
What is the reference plane for the ILS glideslope angle?
Is a 3-degree glideslope based from the runway surface slope at the point where the antenna is installed or is it referenced to a zero-degree plane?
If you know of a design document or FAR/AIM reference, please include that as well.

The runway slope doesn't matter, it's 3 degrees below horizontal. Don't have a reference for you, but that's what you see in the aircraft - reference is the horizon.

I'll be surprised if the answer is that it is based in any way on the runway surface slope. Allowing the surface slope where the antenna is installed to determine the glideslope angle would be introducing varying descent rates for different ILSes, and pilots expect the same general descent rate on glideslopes for the the aircraft landing configuration, including power settings. A 2 degree upslope would mean a 1 degree glideslope, lots of power. A 2 degree downslope, a 5 degree glideslope, extremely hard to follow.

It is referenced to the zero-degree plane, and depending on obstacle clearance it may be a little steeper (3.5° or 4°). There is no reason to refer it to the runway slope as it defines a sink rate to speed ratio during final approach.

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text:
Why aren't large, low-speed propellers widely used?
I am trying to calculate different propellers for the simple case like helicopter moving slow up like 1 m/s or staying/hanging in one place.
From generic physics point of view it is more efficient to push back more mass with low speed, then to push back less mass with higher speed.
Efficiency (Kg/Watt) I mean how many thrust in kg I will get from 1 Watt of power.
I did calculation using Prop selector software as shown in screenshot below big propeller 4 meters with only 32 RPM procude 1kg of thrust and power is just 13 Watts. Speed of the end of propeller will be 6.7 m/s.
Small propeller 27 cm with 7000RPM with the same 1 kg thrust needs 176 Watts of power. Speed of the end of propeller will be 99 m/s. 13 times more power!
Red thrust in the screenshot means "propeller is stalled", but maybe this "stalled" does not make sence for static case, when propeller is not moving.

My question is Why this type of propeller (big diameter, but low speed) is not in use? Is it just practical consideration like big field will be needed if I make very big propeller, or my calculation is wrong or something else?

Propeller stall should mean the blades—which are airfoils like wings—are stalled. That can happen at any forward speed. Probably you need to adjust the pitch angle at the last line—most airfoils stall below 15°, so the 20° and 17° values in your screenshots are way to high for static case. Reducing the pitch angle below stall should increase the efficiency (in both cases).

You are not wrong, it is more efficient to accelerate a large mass by a little than a small mass by a lot.
This is due to momentum being linear with speed and mass, while energy is linear with mass but quadratic with speed, so the same momentum can be obtained more efficiently by slowly pushing a large amount of air, e.g. with a large propeller.
The reasons against this are as you imagined, like clearance from the ground and other parked craft. Also, extremely long propeller blades will suffer from high inertia and bending moments, without the benefit of centrifugal stiffness. And then there is the issue of keeping the propeller tips subsonic, to avoid wasting a lot of energy in the form of sound.

Also there is the parasite drag. Longer blades have more parasite drag and at some point the parasite power becomes more than induced power.

Of course, analogous reasons pro and con apply to long (nonrotating) wings.

That makes sense.  A Bell 47 helicopter gets way over 3000 lbs of thrust from its 37 ft 300 rpm horizontal "propeller" on only 280 hp (with a chunk of that going to the tail rotor), well over double what that engine can produce turning an 80 inch propeller at 2700 rpm the normal way (maybe 13-1500 lbs thrust).  But to use it like a regular propeller would require 20 ft tall gear legs.
So from the standpoint of "as long as possible", setting aside the issue of tip speeds and gear reductions and all that, the prime issue is ground clearance for your prop on a given airframe.  The Osprey gets away with those fabulously long rotor/propellers that are super efficient for cruise because it turns them up to hover for landing and doesn't have to deal with the ground clearance problem.
On a regular plane though, you are stuck with a few of feet of radius to play with, and ideally you use the longest prop that can run with the tips subsonic at your engine's rpm limit, without hitting the ground.

"doesn't have to deal with the ground clearance problem" – Until it has to land with not enough power to go vertical …

I think there is also one very important reason:
Ability to fly/hover in reasonable windy conditions.
When the propeller move air very slowly, even small turbulences can blow the aircraft away.
The extreme case is Human-powered helicopter, it can generate 128 kg lift with only 1.1kW. But they need to do it in indoor stadiums, because any breeze can cause the helicopter lost control.
One more reason is the trade-off between cost/complexity and efficiency.
Especially for small drones, most motors run more efficient at relatively high speed, if you want to use large/slow propellers, you need to use gears to reduce the RPM, thus add weight and cost.

Large propellers are not always  better. It depends on the speed. From momentum theory, efficiency depends on decreasing induced velocity, and thrust is proportional to mass flow. At slow speed, you want a large propeller to increase mass flow and reduce induced velocity, but at high speed mass flow is large and you would want to decrease drag generated by large blades.
As speed increases, mass flow also increases, so the diameter is having less “weight” on the efficiency. As the speed is increased, blade drag “takes over” driving efficiency to drop faster on the larger propeler.

It would be nice to add a bit more information abut what is plotted in the graph.

Don't most of those Reno racer warbirds have rather large props?

Yes, what is the blue and red line means? Also, I mean that the big propeller will be moving at low speed. As in my example in the question, 4 meters propeller with 32 RPM produces the same thrust as 27 cm 7000 RPM. What if I make 500 meters diameter propeller and with only 5 RPM can have 1000 kg thrust?

The calculation with the same program shows, that if I make 1000 meters diameter propeller and with only 0.02 RPM I will get 1200 kg thrust and 1800 Watts power will be needed. I only will need 1.5 watts per kg. for 27 cm propeller and 7000RPM I will need 176 watts per kg.


text:
Does Buoyant Lift Force scale with Load Factor?
Let's say I am flying an airplane that has a weight of 100,000 kg with a 1G load factor at normal steady flight. I have helium lift assist that is providing 50,000 kg (50%) worth of buoyant lift force upwards. Therefore, my aerodynamic lift has to provide 50,000 kg (50%) of upwards lift to remain in flight.
If I go into, let's say, a 60 degree bank maneuver and induce a 2G load factor to the airplane (so now the perceived weight goes up to 200,000 kg), does the buoyant lift force scale in some manner with the load factor, or is the lift force completely static and continue to provide 50,000 kg worth of positive lift regardless of the induced load factor? In other words, does the aerodynamic lift have to provide 150,000 kg of positive lifting force during a 2G maneuver to keep the plane airborne?

As an aside to this hypothetical question, 50,000kg of helium buoyancy is a Zeppelin Graf sized envelope which is unlikely to ever achieve a 60 degree/2g bank.

Pilothead, as human beings, we are vulnerable to confusing the unprecedented with the improbable.

It would probably be easier to think of it as a 50,000kg aircraft with a "normal" wing.  This untruth will make calculating accelerations and turn rate incorrect (because the actual mass will make it more sluggish).  But the wing at all times has to provide 50,000kg of vertical lift for the craft to not accelerate downward.
Then just like a regular plane, the wing has to double the (aerodynamic) lift at 60 degrees.  This allows it to have 50,000kg vertically and the craft doesn't descend.

If I go into, lets say, a 60 degree bank maneuver and induce a 2G load factor to the airplane

I don't prefer this language.  The load factor is the result of the lift.  If you had a wing that couldn't provide increased lift (it was already at the point of stalling), then banking wouldn't increase the load factor.  It would either stall the wing or the plane would start accelerating downward.  Saying the load factor has increased suggests to me that the lift has already been accomplished.

does the buoyant lift force scale in some manner with the load factor, or is the lift force completely static and continue to provide 50,000 kg worth of positive lift regardless of the induced load factor?

The buoyant force is constant and always vertical.  Instead of coming from the relative wind, it comes from the vertical pressure gradient in the atmosphere.  So in the simple analysis, it isn't affected by banking.  But since the gravitational pull on the airplane is also constant and always vertical, you can just combine them to understand what the aerodynamic wing is required to do.

Aerostatic lift is always vertical, and doesn't contribute centripetal force in a turn. If you're in a 2G 60 degee turn, the total force in the vertical direction is equal to the aerostatic buoyancy plus the total aerodynamic lift (pointed 60 degrees from vertical) times cos(60 degrees). The centripetal force is just aerodynamic and is equal to the total aerodynamic lift times sin(60 degrees). To be in a 60 degree 2G turn you would have to be accelerating aggressively upward--for example, turning at 60 degrees while pulling up through the bottom of a dive.
To determine the bank angle required to maintain a 2G level turn with half of the aircraft's weight offset by buyancy, we have two equations and two unknowns:
Unkowns:
theta # Bank angle
lift # Expressed as acceleration, pointed theta radians from vertical
Equations:
lift*cos(theta) + 0.5G of aerostatic buoyancy = 1G # Required for a level turn
2G = sqrt((lift*cos(theta) + 0.5G)^2 + (lift*sin(theta))^2) # Setting the total acceleration equal to 2G.
Now we solve:
=> lift*cos(theta) = 0.5G
=> (lift*cos(theta) + 0.5G)^2 = 1G
=> 2G = sqrt(1G + (lift*sin(theta))^2)
=> 3G = lift^2*sin^2(theta)
=> 0.5G = 0.289*lift*sin(theta)
=> lift*cos(theta) = 0.289*lift*sin(theta)
=> cos(theta) = 0.289*sin(theta)
=> theta = atan(0.289)
=> theta = 1.29
=> lift*cos(theta) = 0.5G
=> lift*cos(1.29) = 0.5G
=> lift = 1.803G
1.29 radians is the same as as a 74 degree bank angle.
At 60 degrees in a level turn...
0.5G = lift*cos(60 degrees)
=> lift = 1G
Total G load = sqrt((1G*cos(60 degrees) + 0.5G)^2 + (1G*sin(60 degrees))^2)
Total G load = 1.32G
At 60 degrees and 2G total acceleration...
2G = sqrt((lift*cos(60 degrees) + 0.5G)^2 + (lift*sin(60 degrees))^2)
=> 4G = (lift*cos(60 degrees) + 0.5G)^2 + (lift*sin(60 degrees))^2
=> lift = 1.703G
Upward acceleration = 1.703*cos(60 degrees) + 0.5G
Upward acceleration = 2.203G
Net upward acceleration = 1.203G.

60 degree turn is a standard training maneuver and is accomplished by rolling the plane into a 60 degree bank (you now have 0.5 g vertical lift)

Pulling the elevator to 2g now gives you 1 g of lift.  On the real world, one can also feed in some rudder, pointing the nose skyward.  That, along with the upward pointing prop, also helps vertical lift.  The elevator is now also turning the plane through the circle.  One can control altitude by adding power, or rolling to a steeper or shallower bank.

Critical here is sufficient airspeed to avoid accelerated stall but not to exceed the structural limits of the aircraft.

@RobertDiGiovanni thanks for the comments. The aerodynamic "lift" in my answer is actually the net aerodynamic force, so perhaps it would have been clearer to call it that. In other words, it is the combination of drag, thrust, and aerodynamic forces from the wings, fuselage, and empennage. What do you think could be improved in my answer?

Certainly nothing wrong with fine tuning a response.  Especially after reading and learning from others on this fine forum.  After further consideration, I would divide my answer into actual lift required in the light and heavy case, and also the "percieved" weight which may translate into structural loading.

Interesting here is a "heavy" might try for 2G at a higher airspeed.  If plane flies at 50 knot, 70 (squared) will double lift, 100 will quadruple.  May wish to factor in speed, as long as it is safe.

Short answer is “no”. The buoyant force is due to a completely different principle, and is always the same amount, pointing up. No scaling.
By the way, for your partially buoyant aircraft, if you wanted to do a 2G level turn, you would need more than 60 degree AOB.
A lot of times, if you think things through to an extreme, it will help you understand the in-between cases.
Imagine, your airplane has so much helium that it is neutrally buoyant (or is flying in a fluid that is dense enough...think submarine). Then what angle of bank would you need for a level turn? You’d need a 90 degree AOB for any level turn.

You may have answered your own question by saying
"percieved" weight.  Note that a 2G turn does not increase the weight of the plane there for does not
increase the amount of VERTICAL lift required to hold
it in the air. 2G is what you perceive or feel in the
plane.  The amount of vertical lift required will always be the weight of the aircraft minus the lifting
gas contribution. The remainder must be provided by
the wing/engine.
So what must be determined is whether or not your plane is capable of producing enough total lift to generate 50,000 kg of vertical lift at a 60 degree
bank.  This would be however close to your max Clift
you dared to venture before stalling.  Another solution would be to use more helium, or reduce the
angle of your turn.  For recreational flying, trying to hold altitude in a sustained 60 degree turn may not
be worth pursuing for long.  You may be spilling fuel
from your wings.
If you still want to do it, remove the helium and see
if the wings will lift 100,000 kg on their own in level flight.  At 60 degree bank, if my math is correct, will give 50,000 kg vertical lift.
Bob
As a post script, this question is applicable to comparing weight vs stall speed. The "light" plane needs less AOA than the "heavy" at a given speed.
There for, the heavy must either pull more AOA to hold
altitude (risking stall) or increase speed (limited by safe maneuvering speed).
What this translates into is the "light" would have less drag, which would save fuel on a long trip.  One might think of filling a Super Guppy with helium to find out!


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What is the regular stall speed of an F-16 flying in a straight-and-level path?
I want to know what is the minimum speed that an F-16 can fly at without stalling.
I have searched the web, but found no exact information answering my question.
Most of them say that it depends on many factors, like payload and maneuvering. But, for simplicity, I want to know the stall speed of that plane flying in a straight path at a constant altitude.
If there is no exact answer, I want to know the regular stall speed, just an average value of that speed, for a clean F-16A weighing 18,000 kg flying in a straight & level path.

Comments are not for extended discussion; this conversation has been moved to chat.

I don't have exact data either, but we can easily make an estimate with data from Wikipedia. Note that I explicitly neglect high-lift devices which are present on the F-16 (notably the strakes which generate significant vortex lift), the variable angle of incidence (washout), as well as body lift which probably plays a significant role as well.
I estimate the aircraft weight to be around $150\,\mathrm{kN}$. The wing area is approximately $28\,\mathrm{m}^2$, and the NACA 64A204 airfoil used has a max $C_L$ of 0.8. Fill that in the lift equation $$L=\frac{1}{2}C_L\rho v^2$$
we get a speed of about $100\,\mathrm{m/s}$ which is about 200kts.
At the angle of attack of max $C_L$, the L/D is about 10, so the engine has to deliver $15\,\mathrm{kN}$ of thrust to maintain level flight. This angle of attack is approximately 6 degrees, and due to the probably positive angle of incidence between the fuselage and wings, the thrust angle is likely less than 6 degrees w.r.t. the horizon. This means that the vertical component of thrust is about $\sin(6°)\approx 0.1$ times smaller, or $1.5\,kN$. Given the error bars on my estimates (especially the aircraft weight), this does not warrant a recalculation because most of the weight is indeed carried by the wings.

200 kts seems like a really fast stall speed, yes?  Aren't most planes around 180kts on approach, and land even slower than that, yet are still above their stall speed?

@abelenky I agree. Then again, this is a high-performance airplane, and I neglected high-lift devices which significantly delay stall.

I does not want the speed to be accurate but an approximated number. just a number. Any one have tested this or near this stall speed, this will be helpful

@Ahmed If you read my answer more carefully, you'll notice that the answer is highly dependent on weight. This can change significantly based on fuel and weapons carried. There's no "number", just different numbers for different weights, following roughly the relation given in this answer.

The F-16 is not a glider, it uses vortex lift from the strakes to lower its minimum speed. Without wind tunnel data it is very hard to calculate its maximum lift coefficient, but I would guess it is around 1.1 using the wing for the reference area and adding tail lift on top of wing and fuselage lift. 85 m/s looks about right to me.

It depends on the weight of the F-16. If weight is below afterburner thrust the F-16 can fly vertical, without having to worry about any lift that the wings are generating. At airshows it passes by straight and level with the nose pointing almost straight up.
So its stall speed in that situation is Not Defined: even at speed zero there is no stall.

I think the "straight flight path" + "constant altitude" part of the question removes the afterburner vertical climb solution.

Not vertical climb, the video shows a constant altitude, horizontal flight path with the nose pointing almost straight up, the plane being supported by thrust only and streamflow over the wing being irrelevant.

How does the F16 maintain control authority in this position?

That video clip you've posted (0.53) is somewhat misleading: it's a radio-controlled model.


text:
How should Isopropyl Alcohol be used as an AvGas Fuel System Icing Inhibitor?
I understand that 100% Isopropyl Alcohol (IPA) is used as a Fuel System Icing Inhibitor (FSII) additive in aviation gasoline systems—including 87 through 100LL fuel types—for cold weather operations. I know that IPA is, or can be, included as part of AvGas formulation. However, I am referring to operational use of IPA as an additive when fueling the aircraft. Pure IPA is also known as anhydrous isopropanol.

What documentation is available to support such use?
What is a proper treatment ratio for using IPA as an FSII (or what is the concentration limit)?

FAA Advisory Circular AC No: 23.1521-1B deals with use of IPA for Part 23 aircraft. It states:

ASTM D 910, Standard Specification for Aviation Gasolines, allows the use of isopropyl alcohol conforming to the requirements of ASTM D 4171, specifications for Fuel System Icing Inhibitor, as a fuel system icing inhibitor. Accordingly, isopropyl alcohol conforming to ASTM D 4171 may be used in concentrations up to 1 percent by volume, to benefit safety, as an icing inhibitor in automobile gasoline.

However, it is left to the aircraft manufacturers (according to ASTM 910 too) to determine whether or not to use IPA as an additive. For example, in a letter regarding Lycoming engines in certain of their aircraft, Cessna says,

Isopropyl alcohol in amounts not to exceed 1% by volume can be added only to aviation fuel  (not  automotive  fuel)  to  prevent  ice  formation  in  fuel  lines  and  tanks.  Although approved  for  use  in  Lycoming  engines,  do  not  use  isopropyl  alcohol  in  the  aircraft  fuel systems unless approved by the aircraft manufacturer.

Note that Cessna allows the addition of IPA only to aviation fuel for the applicable aircraft. In UK, the MoD Defense Standard 91-90 again leaves the matter to the individual manufacturers, saying,

The concentration of Isopropyl Alcohol shall be recommended by the aircraft
  manufacturer at the time of delivery to the purchaser and typically not exceeding 1% v/v.

Though almost all these cap the IPA at 1%, the best way is to refer to the individual aircraft's appropriate manual as it would have the necessary details. But generally, I think that manufacturers (of large aircraft, atleast) prefer not using IPA these days due to environmental concerns.

I wonder what the effect of a Peterson Auto Fuel STC is on Cessna's note regarding automotive fuel.

It seems as though there was an accident involving a PA-23-180 a while back that lead to this advisory. You can see the full report here (interestingly this was published by the NTSB not the FAA).

Require that Piper, Beech, and other airplane manufacturers who have
  not already done so issue service and operating information regarding
  the use of fuel additives in piston-powered airplanes for cold weather
  operation and incorporate this information in the pilot operating
  handbooks of all newly manufactured airplanes. (Class 11, Priority
  Action) (A-85-79)

In terms of limits it looks like it might be in the POH and should be in the AFM for the aircraft or in a service bulletin like this one issued for MD helicopters.
The AFM for this 172S (page 4-23) notes the use of IPA and redirects to section 8 (page 8-17) which says a 1% by volume quantity should be used. It also offers some steps on how to add it to the fuel and a nice chart on volume of additive vs volume of fuel.

I fly a PA-30 Twin Comanche regularly,and,since the POH does not discuss use of isopropyl alcohol or Prist,have contacted Piper repeatedly to seek approval.No response has come from Piper.AOPA and Int’l Comanche Society are equally silent on this topic.I even requested that Air Safety Foundation look into it.
I consider this issue of great importance, as the PA-30 fuel system lacks individual drains for main & aux tanks, putting it at high risk for ice formation in its fuel system.This system, along with its suboptimal procedure of draining the fuel bowl via the selector valve was the subject of criticism by NTSB in reviewing a Piper Apache fatal accident blamed on icing and obstruction of the fuel bowl.This led to a Safety Recommendation, apparently ignored by Piper.Beech & Cessna both had statements of approval in POH for use of IPA or Prist for my previous A36 & C310 aircraft.
Why is everyone so mute on this subject?
Piper ought to be held accountable as irresponsible if preventable fuel system icing causes a future accident.

The answer by Dave indicates that Piper should rectify this, if they haven't, I'd suggest posting a new question about how to address that issue as it will get you the best exposure and most advice.

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text:
What constitute "known icing conditions" and when can you fly VFR while they exist?
For flight operations conducted under 14 CFR part 91(other than subpart F) in VMC, what conditions constitute "known icing conditions"? Under what circumstances (e.g., altitude difference or no visible moisture), if any, could a VFR flight operation (other than in an aircraft certified for flight into known icing) be legally conducted despite an Area Forecast, AIRMET Zulu, Pirep or other possible indicator of known icing conditions?

I was always told that airframe icing will only occur when flying into visible moisture. Maybe my flight instructor was wrong, but he seemed to be pretty smart...

@Ron, I think that's true, but VMC can include flight through varying degrees of visible moisture.

Visible moisture in the air is a cloud, rain, or snow, on the ground its fog. VMC you need to remain free of clouds in all VFR conditions, the only questionable one I think is ground fog where visibility is still VFR.

@Ron, visible moisture can also include precipitation, which can cause airframe icing.

Right, I edited my comment probably before you saw it, my point is to dispute this in your post: "Under what circumstances (e.g., altitude difference or no visible moisture)", I'm pretty sure you can't have airframe icing without visible moisture.

@Ron, OK, right. The reason I asked the question and included that clause is that certain information available on the subject would seem to suggest that any AIRMET Z, etc, could constitute known icing conditions, regardless of the presence of visible moisture along the actual flight path.

This came up recently in another ASE Q or A. It has to do with a specific, FAA-prescribed temperature range that is actually extremely broad and at its outer edges, is rather unlikely to actually create icing.

The short answer is that it's up to you as PIC to make that determination; the FAA has no clear definition apart from what a "reasonable and prudent" pilot would do.
The FAA issued a legal interpretation on exactly this question in 2009, here's the closest thing to a definition that they give (my emphasis throughout):

"Known icing conditions" involve instead circumstances where a
reasonable pilot would expect a substantial likelihood of ice
formation on the aircraft based upon all information available to that
pilot. While "known icing conditions" are not defined by regulation,
the term has been used in legal proceedings involving violations of
FAA safety regulations that relate to in- flight icing. The National
Transportation Safety Board (NTSB) has held on a number of occasions
that known icing conditions exist when a pilot knows or reasonably
should know about weather reports in which icing conditions are
reported or forecast

They go on to say that the PIC is responsible for determining the likelihood of icing conditions:

Any assessment of known icing conditions is necessarily fact-specific.
Permutations on the type, combination, and strength of meteorological
elements that signify or negate the presence of known icing conditions
are too numerous to describe exhaustively in this letter. Whether a
pilot has operated into known icing conditions contrary to any
limitation will depend upon the total information available to the
pilot, and his or her proper analysis of that information in
evaluating the risk of encountering known icing conditions during a
particular operation. The pilot should consider factors such as the
route of flight, flight altitude, and time of flight when making such
an evaluation

Note that even visible moisture and freezing temps aren't automatically considered to be known icing conditions:

The FAA does not necessarily consider the mere presence of clouds
(which may only contain ice crystals) or other forms of visible
moisture at temperatures at or below freezing to be conducive to the
formation of known ice or to constitute known icing conditions.

The specific regulations mentioned in the interpretation are 14 CFR 91.9 (compliance with the operating manual), 91.13 (no careless or reckless operation) and 91.103 (become familiar with all information relevant to the flight). Those regulations would be the basis for any hypothetical enforcement action:

If the composite information indicates to a reasonable and prudent
pilot that he or she will be operating the aircraft under conditions
that will cause ice to adhere to the aircraft along the proposed route
and altitude of flight, then known icing conditions likely exist. If
the pilot operates the aircraft in known icing conditions contrary to
the requirements of § 91.9(a), the FAA may take enforcement action.

So the bottom line is: you're PIC and it's your decision whether it's safe to fly or not because there are far too many variables for anyone else to decide. But personally I think it would be difficult for any pilot to say that it was "reasonable and prudent" to fly a non-FIKI aircraft in an area where forecasts, AIRMETs or PIREPS indicated icing.

The last sentence of your question captures the sentiment that led me to ask the question in the first place. Would you consider that it was "reasonable and prudent" to fly a non-FIKI aircraft in an area where forecasts, AIRMETs or PIREPS indicated icing, if you maintained VMC and avoided areas of precipitation that caused ice accretion?

What is reasonable depends entirely on what the pilot observes. All the legal interpretations, forecasts, and even pireps count for nothing if your observation of the actual weather conditions at the exact time and altitude you penetrate an area of forecast icing indicate to you that it's safe. This FAA interpretation only simplifies enforcement action after a plane turns into an ice block. As far as I can tell, it has no practical value other than helping the FAA build a case for a violation.

@JonathanWalters Maybe, but it all comes down to the details. How high are the ceilings? Where's the freezing level? Are there any fronts moving around? Do I know the route and aircraft well? Etc. At one extreme I'd certainly do pattern work, but at the other I wouldn't set off on a 250nm night XC. In the end this is all about risk, experience and judgement: I might refuse to fly in conditions that an Alaskan pilot wouldn't hesitate to fly in. And we might both be 'correct' about our decision.

The link to the legal interpretation is dead.

Re "The link to the legal interpretation is dead" -- it works now, and any answer containing a link to an actual interpretation of this nature automatically gets an upvote from me-- thanks for fixing J Walters--

I applaud the FAA for attempting to add clarity to a difficult subject.
One of the points in the FAA’s legal interpretation I took, was the importance of the pilot determining an “Exit Strategy” in their flight planning: what to do should the pilot encounter airframe icing in-flight.  This approach is similar to the way we are all taught to look for a field for an emergency landing in case of a loss of an engine (i.e., SEL Airplanes).
The “reasonable and prudent” is the part that concerns me. What is “reasonable and prudent” to a pilot with years of experience may not be to a pilot with just a few years of experience. Perhaps the FAA considers “reasonable and prudent” in the context of other pilots with the same experience level, but that is not how I read the letter.
Legal Interpretations can be found below in the event the AOPA website link breaks.
https://drs.faa.gov/browse/LEGAL_INTERPRETATIONS/doctypeDetails

I'm not seeing that this provides an answer to the stated question.

I think you might be interpreting “reasonable and prudent” as if it is a subjective non-entity, i.e. based mostly on opinion which could vary depending on the individual and experience.  The concept is actually more one of a fixed reality which requires judgment to ascertain. Pilots are expected to begin developing various types of judgment pretty early in the training process. With experience judgement can improve, but it's an improvement in recognizing the reasonable and prudent decision, it doesn't change what the reasonable and prudent decision is.

The question "Under what circumstances (e.g., altitude difference or no visible moisture), if any, could a VFR flight operation... " The answer "Have an Exit Strategy".

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What are ISA +10, ISA -15 and ISA -30?
I have found out that an ISA atmosphere is:

15 °C temperature
5 °C dewpoint
1013.25 hPa pressure

What are the conditions of an ISA +10, ISA -15 and ISA -30? I know the temperature, but what are pressure and dewpoint?
I need to create those conditions to do some flight testing in a simulator.

Related: How is pressure related to air density? (but doesn't address dew point)

ISA+X is the ISA atmosphere with X°C of variation.
It´s used to determine how diferent (in terms of Temperature) of the Standard the actual condictions are and adjust the calculations.
ISA as you said is:
+15°C / 0ft /  1013.2 hpa / LAT 45° / Term. Grad. -2°C/1000ft
So, ISA +5 is:
+20°C / 0ft / 1013.2 hpa / LAT 45° / Term. Grad. -2°C/1000ft

The pratical use is like the following:
If you are at FL075 and the OAT is -15°C whats your difference to ISA?
ISA Air Temp. @ FL075 is:
ISAt = 15 - 2 * alt/1000
ISAt = 15 - 2 * 7500/1000
ISAt = 15 - 2 * 7.5
ISAt = 15 - 15
ISAt = 0°C

So you are 15°C below the ISA temp at your FL or simply ISA-15.

To be exact, lapse rate is 1,98 °C / 1000ft.

So only temperature varies and not sealevel pressure?

Yes, Sami, i just rounded it up for easy calculations.

@Invariant the ISA+X is the deviation of temperature to the ISA Ceteris Paribus.
This is because while Flying QNE you "simulate" flying ISA assuming that the Pressure is 1013.25hpa

@Invariant The sea level pressure in ISA is always 1013,25hPa. When the atmosphere temperature changes (ie. ISA+25) also the pressure on other altitudes changes. Check the pdf on my answer for the formula.

@AndreasAlmeida QNE is not an altimeter setting. QNE means "threshold altitude using standard pressure setting".

First of all:

I have found out that an ISA atmosphere is:
   15 degC temperature 05 degC dewpoint 1013.25 pressure

Not entirely correct. The temperature and pressure are ISA values at mean sea level. These alter with altitude.
Dewpoint is not defined in ISA, as ISA does not contain dust, moisture and water vapor.
ISA-30 atmosphere means temperature deviation from ISA temperature. This means that on a given altitude, temperature is 30 degrees colder than in ISA. Respectively ISA+10 states 10 degrees warmer.
Below is a table of ISA values.

For more physics for calculations, here is a very nice pdf on the matter.


text:
Does intake mass-flow rate vary linear proportional with turbine fan speed?
Is jet turbine intake mass-flow rate linear proportional with its fan speed (N1)?

Or is maybe more linear related to fuel consumption rate, or thrust?

Is this a homework question? It kind of sounds like one.

Mass flow and thrust are closely related, and we have answered how thrust depends on engine speed here. The short answer is No.

No its not homework.
Thank you, @PeterKämpf

Oops, I meant fan speed, so N1, editing the question.

Could you explain what N1 is, for lay readers?

Done now @daniel

Not linear, because the back side of N1 will influence the flow. Not only that but the flow is non linear with respect to rotation in most axial flow compressors.
Is there a specific engine or cycle that you are wondering about, because that changes the answer to your question?

Yes, the engine is Volvo RM8. I am simulating it, and have some info on it when its running, I need to find something that is linear or close to that with regards to the intake mass flow. I have thrust, fuel consumption, N1 and N2.

Then I would stick by the non linear model, as it is a two spool turbine.  If you can't get good data form airframe applications, that was used for ship propulsion, and you might get data that way. Remember that N1 is quadratic to thrust.

You might look up a paper by T Abbott of Langley (1990) on engine monitoring.

Many thanks @mongo.


text:
Turbo-fan idle at high speeds, will air speed up fan speed?
If N1 idle is 30%, will flying idle at very high speeds force the fan to go higher than 30%? (for example idle while descending fast)
And if no, why not?

yes it will, because then it will create less drag.

When an aircraft is descending at idle, some engine controls actually keep the idle speed (rpm) high, at least in military aircraft.

Yes, N1 should increase as the plane reaches higher speeds. This means that the blade speed will increase, therefore, increasing N1. Now because the engine setting is constant (throttle), this means that only a slight increase in the amount of thrust will happen.

Why does N1 increase with increasing airspeed?

N2 doesn't increase does it?

@Koyovis you can refer to this to see what N1 is: https://aviation.stackexchange.com/questions/14690/what-are-n1-and-n2
It's basically self-explanatory because as the airspeed is greater, the fan will speed up.

You can read the post above as well @Invariant . The N2 will increase as it is once again, the speed of the blades in the high-pressure shaft whereas N1 is the speed in the low-pressure. Either way, both should increase.

@SethDeegan I know what N1 stands for, that wasn't my question.

Did I answer your question? Sorry for the extra explanation @Koyovis .


text:
How much does it cost (on average) to run an airliner as a low cost airline on short distance flights?
Considering Germany is thinking about implementing new taxes to increase costs for commercial flights due to how cheap flights have become and the pollution they produce, I was thinking how much does running an airliner actually cost.
I've seen flights from Germany to Mallorca for less than 20 € and even if those are just a few seats and with next to no service, I was wondering, how much does a flight like that acutally cost. - Including fuel, all fees, the crew (average wages), maintainance cycles and if possible the aircraft purchase price itself.
For example, if an aircraft does 6 two hour flights a day and cost an initial 108 million € for an A320neo, assuming a 30 years lifetime and a resell value of maybe 30 million after that (is that realistic?) that would average the cost of the aircraft itself at roughly 7200 € per day or 1200 € per flight.
Regardless of airlines themselves, what are the ballpark values for all costs (fuel, crew, aircraft, fees, anything I didn't think about...) of an airliner per flight (for the sake of the argument assuming A320neo on short distance fligths, e.g. 2hrs)?

That question was rather for long distance flights and I was thinking more about the short distance flights that acutally led politics to think about new taxes because these seem to get cheaper and cheaper. - My neighbours even took a flight to Morroco for what was basically an "afternoon coffee", just because it only cost them 28€ (including return flight)!

You can still scale the answer to that question to your needs. The fuel burn for an A320 is easily found online; amortization you already calculated; crew cost will be somewhat cheaper on short hauls, but not much; the rest is the same.

Airlines go by cost per available seat mile with wide variations; here is data for the US.https://www.statista.com/statistics/527881/us-domestic-cost-per-available-seat-mile-by-airline/   Taxing success purely to punish it seems to be a counterproductive strategy for promoting public welfare.


text:
Is there a way to code Minimum Enroute Altitude using ARINC 424 file format?
I am studying ARINC 424 file format, and I have noticed that there is not any reserved space for coding Minimum Enroute Altitude (MEA), as far as Instrument Flight Procedures are concerned.
I find this value to be really important for properly specifying the characteristics of a procedure. It seems strange that in ARINC 424 specification, MEA is omitted. I have also checked FAA's Coded Instrument Flight Procedures file (CIFP), which are based on ARINC 424, and MEAs are nowhere to be found.
Perhaps I am missing something? Could someone point me in the right direction?

It's in the Altitude/Minimum Altitude data field:

5.30 Altitude/Minimum Altitude
Definition/Description: The Altitude/Minimum Altitude field indicates
  the reference altitude associated with (1) Enroute Airways (MEA, MFA
  or other minimum altitudes as defined by source), (2) holding pattern
  path of Holding Pattern record, (3) altitudes at fixes in terminal
  procedures and terminal procedure path termination defined by the Path
  Terminator in the Airport or Heliport SID/STAR/Approach Record and (4)
  lowest altitude of the blocked altitudes for a Preferred Route.
  ...
  On Airport and Heliport SID, STAR and Approach Route records, the
  first Altitude field will contain an altitude when Altitude
  Description field contains a plus (+), a minus (-), or one of the
  following characters: B, G, or V. The second Altitude field will
  contain an altitude when the Altitude Description field contains one
  of the following characters: B, C, D, G, or V. In approach procedure
  coding, some fix Altitudes may be below sea level, in the case of
  altitudes at runway fixes when the runway threshold elevation is below
  sea level. In these cases, the Altitude will be expressed in feet with
  a minus (-) sign in the first character of the five-character field,
  see examples.

In 4.1.9.1 SID/STAR/Approach Primary Records you will find columns 85-89 and 90-94 are the two altitude fields referenced in 5.30.
The characters referenced in 5.30 are defined in:

5.29 Altitude Description (ALT DESC)
Definition/Description: The Altitude Description field will designate
  whether a waypoint should be crossed at, at or above, at or below or
  at or above to at or below specified altitudes. The field is also used
  to designate recommended altitudes and cases where two distinct
  altitudes are provided at a single fix.

Thank you Gerry for the provided quotes from the ARINC-424's specification manual. However, isn't the aforementioned field used to describe the altitude constraints imposed on a waypoint (e.g. "CROSS LARRY AT 2500 OR ABOVE"), as far as flight procedures are concerned? Isn't this different from MEA, which applies to the actual segment?

This supports both concepts. All data is tied to waypoints; e.g., an airway is defined by a list of waypoints. As stated in the first definition in 5.30 "...(1) Enroute Airways (MEA..."  You should also note the 5.29 descriptor where it addresses  "cases where two distinct altitudes are provided at a single fix."  Thus it can define the MEA as the minimum crossing altitude based on the direction crossing the fix.

Thank you, I really appreciate your thoughtful answers. I understand it more clearly now, however I wonder what will happen if there are both an altitude constraint at a specific waypoint and minimum enroute altitudes assigned to its adjacent legs? Take HAWKZ7, where one can see PTERA "AT/ABOVE FL220" but at the same time MEA "10000" assigned to both adjacent legs (and additionally MOCA values, not even mentioned in ARINC manual...). Altitude data field would be populated by FL220. Which field MEA would populate?

First, your link it HAWKZ7 is incorrect.  Having found it, I can see your question.  The issue is that you're looking at the 'altitude field' for PTERA as it is used in the STAR.  The 10000 ft MEA is a moot issue for the STAR.  You would need to find the Airway record for PTERA.  The airway definition, not part of a procedure, should have the actual MEA listed in the altitude field.

Once again, I am grateful for the thorough answer and I apologize for the inconvenience concerning the wrong link. The last piece of the puzzle is how one can make the connection with the related airway, in order to get MEA. For example, in FAA's CIFP database, PTERA is a terminal waypoint, defined for KSEA airport, not an enroute waypoint. Thus, it is not mentioned in any airway records. Is there any way to find the airway with the MEAs mentioned in HAWKZ7's chart?

I don't have an answer to that. I'm a database user, not a developer.  I suspect what you're looking for is in the Aeronav 'master' database.  As a terminal waypoint, the MEA is a factor for designing the procedures, so it obviously exists.  Since the waypoint isn't intended for use outside of the procedure, the MEA doesn't appear as part of the procedure in the published database as the procedure altitude takes precedence.

As the name suggests, MEA is coded for enroute segments lik ethe airways (ARINC permits this).  In so far as Instrument Approaches are concerned, the templates takes into account vertical clearances in Primary and secondary areas and then arrives at safe heights at each segment. Geometry of primary and secondary areas and the clearance therein depends on types of approach. For example, in ILS approach, half full scale deviation is considered safe. Also, it stems from this, obstacle clearance at different segments of the approach would be differen, so, specifying a minimum altitude may not be applicable. Secondly, in emergency, Minimum Sector altitude (MSA) centered 25 nm around airport/nav aid provides requisite obstacle clearance..

This comment does not answer the question.

Thank you Sandeep for your contribution and welcome to aviation stack exchange!  Although you have made some valid points concerning the definition of Minimum Enroute Altitude, still in most SID, STAR or IAP procedures one can clearly see that MEA values are important. Take, for example, this STAR, this SID and this IAP from JFK Airport. Even intermediate segment has a coded MEA (1900).


text:
Why do localizer/glideslopes use 90/150Hz specifically?
Why is it that localizer and glideslope system are using 90 Hz and 150 Hz frequencies only, instead of some other LF such as 75 Hz?
Part two of my question:
How does the modulation depth differ for aircraft using localizer and GS receiver?

Do you have any reason in mind why they would use other modulation frequencies besides 90 and 150 Hz?

Because if they used 75 Hz people would ask why they don't use 90.

An AC motor in the USA will turn at 60 RPM and 90 is 3/2 and 150 is 5/2 of 60. Both can be generated using a simple gear drive which would reduce the costs during development. Similarly there are circuits to multiply or divide a driving frequency by whole numbers, with 60 hertz readily obtained. Simple gear ratios could convert 50 hertz used elsewhere.
90 is two digits and 150 is three digits. Left is four characters and Right is five characters which makes for a handy memory aid, as the greater number of digits matches the longer word. Similarly 180 hertz is commonly used to spin up X-ray tubes as a 60 rpm motor turns a generator via a 3:1 gear ratio. Analog clocks are kept accurate by exact 60 hz power and is better regulated than that of a common occilator, especially during development. Also the developers probably didn’t want to use 60 hz or a Harmonic of it to reduce noise from the power lines.

Javelin thanks but your source?

Add to that the 1020 Hz used for the ident signal of the localizer, 17x60.

The frequencies are chosen to allow the system to produce a localizer and glideslope 'beam' that provides proportional guidance that can adequately guide an aircraft on the approach.
Read through this Question and Answer.
As you can see in the description of the beam-forming, the width of the antenna array and the choice of the modulation signals (90 and 150 Hz) drive the ddm proportionally to the angle of approach to the antenna array.
There are also considerations tied to the broadcast signal. Higher frequencies would require more bandwidth which would impact channel spacing.
Ultimately, the choice of the two frequencies is an optimization of a design for the approach guidance. For any given runway the current system can be set to provide consistent guidance (+/- 350 feet == full scale deflection) at the runway threshold.  This is done by adjusting the distance between elements of the antenna array.

Thnx Gerry that seems interesting and logical answer however i have different theory !!! Back in old days to generate electricity u need Alternator( turbines) that run at frequency of 60Hz. So to choose anything which is multiple of 60Hz would be easy but if we use e.g 60Hz or 120 Hz in ILS, will cause interference so chose frequencies of 90Hz and 150Hz which were multiple of 30Hz and remember 30Hz was fundamentsl LF frequency easily developed out of 60Hz. I hope it makes sense :)

@aadeez  That is a valid assumption but it's not just 'back in the old days'. Any system that has to generate an AC signals has to consider how to most effective do so.  Almost all use a single reference, whether an internal oscillator or an external signal. Using the ground power 60Hz as a reference would simplify making 90Hz and 150Hz signals as they are 1 1/2 times and 2 1/2 times the reference. But that's not the only reason, just one of many considerations. In the end, somebody ran through the trade-offs and picked these as the best choice.

Agreed Gerry :) i posted one more query of why localiser use VHF and glideslope, UHF? still to be answered by some one. Any thoughts?

I think he wants to know why they chose specifically 90/150 Hz TONES on the radio signal, not the radio signal freq itself. I'll answer below.

Never mind, Gerry's answer is technically correct (bandwidth considerations). Also they are not harmonically related. The envelopes cancel out via the antenna radiation pattern such that 90Hz cancels out to the right of the course and below the glide slope, and vice versa for the 150Hz. Very clever actually. It has certainly stood the test of time. Back in the 1970s the FAA attempted to replace ILS with Microwave Landing System. It failed.

@Rumi, I doubt they'd want to depend on the mains power having exactly 60 Hz to the required precision, and even less so for any backup power source. The mains frequency is pretty stable these days as many powerplants have to run in phase, but I doubt it was the case in 1930s.

When ILS was developed, they actually used rotating capacitive plates with lobes. The 90 had 3 lobes and the 150 had 5. Easy way to modulate the RF with the phases locked. The depth of modulation was adjusted with a fixed plate with a variable gap between it and the rotating plate.

Answer to part 1: Radio frequencies are split up for different uses by the government (NTIA). The aviation world makes use of many of these ranges, including the range around 75 MHz, but most usable frequencies are already allocated for other purposes. This chart provides a great breakdown of the US allocation of radio frequencies.

Hi Geoff well thanx but its still unclear ! You mentioned about 75MHz but i am talking about Low frequency modulating signals of 90Hz and 150 Hz, not MHZ mate :) Also the Chart that u shared describes as low as 3Khz only.... my point is Why only 90Hz and 150Hz? we could have used 45Hz too :) there must be some logic i guess


text:
Why does the glideslope use UHF when the localiser uses VHF band?
Why does the glideslope use UHF, while the localiser uses the VHF band?  I mean, the glideslope could have used the VHF band too, or vice versa, so why is it so?


text:
What are the procedures for flight checking a glideslope and localizer?
What are the FAA and ICAO requirements and procedures for flight checks and calibration of glideslope and localizer transmitters? I'm interested in both the engineering and piloting aspects.

It's all laid out in Order 8200.1D United States Standard Flight Inspection Manual.
That and other documentation, including the inspection reports can be found from the FAA website under Flight Inspection Information.

Gerry what about ICAO ? And just to be clear i am more interested in knowing how to perform flight check not in reading the procedures only :)

The goal of flight check is to verify the "0 ddm" center of the LOC and G/S signals are propagating in the correct, straight (withing tolerance for bends and noise) path. The flight check a/c flies the approaches using a highly calibrated receiver that logs the data. At the same time a tracking system logs the actual location of the a/c. Older systems used a ground based laser tracker. Newer systems use a precision differential GPS system with a local reference ground station. The data  is the post processed to determine the actual path of the signal in space.

Can you summarize the main points of the doc in your answer? Link-only answers are generally not preferred here because pages move, documents are updated etc.


text:
What is this hole at the front of the engine pylon on a Boeing 707?
Images of the Boeing 707 show a hole at the front of the engine pylon just above the engine.
Can someone please tell me what is the purpose?
Also why do images show the hole absent on the left outer engine pylon but present on engines 2, 3 and 4?

ref: wikimedia commons

This is because those engines have turbocompressors which are used to supply cabin air.
The engine bleed air drives the turbine of the turbocompressor, which in turn would drive a compressor which would pull in air from the atmosphere through the front hole, compress it, and send it down to the Environmental Control System for use as cabin air. The discharge outlet for this bleed air is the long teardrop shaped hole in the side.
The tubocompressor can be seen in the image below.

Norm King at work on the turbo-compressor on Number 3 engine on 17 August 2006. Photo: Karen Glass; image from 707.adastron.com

Wikipedia  tells you the answer:

The 707 uses engine-driven turbocompressors to supply pressurized air for cabin pressurization. On many commercial 707s, the outer port (number 1) engine mount is distinctly different from the other three, as this engine is not fitted with a turbocompressor. Later-model 707s typically had this configuration, although American Airlines had turbocompressors on engines 2 and 3 only. Early 707 models often had turbocompressor fairings on all four engines, but with only two or three compressors installed.[19]

As stated above, the exact configuration varies, so some pictures show all engines the same, or not.

Further info is available from a discussion on a PPRuNe forum:

Turbocompressors were needed because insufficient bleed air was available from the engines at lower RPM's to keep the cabin pressurized during descent. Also, using bleeds reduced the climb performance greatly, whereas the TC's were more efficient at delivering cabin air. In addition, engine bleeds did not supply enough air for sufficient cabin ventilation with a full (189) pax load.
When maximum range was required, switching off all the turbocompressors and using engine bleeds for pressurization gave a 3% improvement in specific fuel consumptioin. This was OK for freighters, but resulted in a rather low cabin ventilation rate with pax aircraft.

Turbo compressor drive was provided by 16th stage air (high pressure, high temperature) from its respective engine. 16th stage air was also used for: engine anti-icing, thrust reversers, fuel heat, hydraulic tank pressure, surge bleed valve and water tank pressure. To operate above 40,000 ft on a B-320 we had to have 3 T/Cs or 2 T/Cs and one bleed or 1 T/C and three engine bleeds or four engine bleeds operating. Penguin is correct in stating that engine bleed air alone was not sufficient in maintaining cabin pressure in a descent. Before descent the flight engineer would start the T/Cs in an attempt to maintain the (8.6 max) cabin pressure differential. Failure to accomplish this would result in an immediate loss of cabin pressure. Normally, we would try to keep the engines spooled-up (bleeds closed above 55% N1) in the descent to augment the T/Cs and reduce the cabin pressure gradually. This procedure was also necessary when operating in icing conditions.

So the asymmetric setup with T/Cs on 2,3,4 but not on 1 is because you only need 3, not 4?  And it's an on/off thing (and 4 would be too many), or there's some other reason it wouldn't make sense to have all 4 engines fitted with a T/C?

The key to answering your question is the weight of the TC.The first B-707 (N-70700) and early models that followed had no turbo compressors installed; same for the Air Force A & B KC-135s. These aircraft were powered by the Pratt & Whitney JT3C-6 turbo jet engine which only developed 11,200 pounds of thrust but could manage 13,500 pounds of thrust with the use of water injection. It took almost two miles of runway to get one of these fully loaded aircraft (MTOW 257,000lbs) to get air born with water injection and the take-off was always interesting. When the P&W JT-3B engine became available delivering 18,000 lbs. of thrust or 19,000 lbs. with the dash seven engine, the MTOW was increased to 326,000 lbs and the weight of the TCs could be more easily accommodated. Some B-320 aircraft did have four TCs and I believe President Kennedy"s Air Force one was so equipped. The third TC was commonly removed from freight B-707s ( #4 engine) to lower the empty weight and allow more cargo to be loaded.


text:
Are fixed wing aircraft with gimbal thrust feasible?
Elon Musk has suggested that he'd look at implementing VTOL electric aircraft with a "gimballed thrust" system similar to the one used in rockets.

Aircraft have all these unnecessary things like tails and rudders and elevators - not needed. Just gimbal.. anyway.. gimbal the electric fan. For some weird reason gimbaling motors is normal in rockets and not in aircraft. Why not? [Do you have specific plans?] I've been sort of toying with the design for an electric supersonic vertical take-off and landing electric aircraft for a while, I'd love to do it, but I think my mind would explode

What are the challenges to implementing this? Is it technically feasible?

That's called thrust vectoring in aviation speak. And doing away with control surfaces makes the propulsion indispensable. You cannot even run out of fuel anymore. Technically easy, but it will be hard to find pilots who will fly this widowmaker.

On a related note: feel free to check out the (many) questions about electric aircraft here, to get a bit of a feel how reasonable a supersonic VTOL electric aircraft is, gimbaled or not. Also related are dirigibles, which can use thrust vectoring, and can provide you with insight what happens when they lose thrust.

@PeterKämpf thanks so much, that term was exactly what I was looking for!

i haven't figured out if Musk is losing his grip on reality or if he just enjoys winding people up and watching them act like idiots. Probably it's both.

Wait - a tail is not needed?

@PeterKämpf isn't he describing an helicopter ?

@Antzi: Not really. In a helicopter the mast stays fixed (apart form rotating, obviously). This is more like the Harrier, only with electric blowers instead of a gas turbine plus a fan.

A helicopter vectors thrust, the blades have 3 axes of rotation.

Musk does explicitly mention "gimbaling the electric fan"

one of the reason why rocket needs it this way is there is a good portion of the flight at low speed and without air, so you need thrust vectoring. Same thing kind of goes with missiles whose flight has either of those characteristics (high altitude or partially very low speed). Planes never fly that high where vectoring is significantly better than surfaces, or need that much maneuverability at low speed. One use case, however, is navy aircrafts where improved low speed maneuverability improves landing safety a lot, or high AoA maneuver where most control surfaces would stall.

@PeterKämpf Well, what about missiles? (Aircraft, aren't they?)

@user3528438: That would be electric missiles, then. Haven't seen one yet.

Could you create an aircraft using an jet engine/propeller/ducted fan on a gimbal?  Sure you 'could'; it is feasible by the laws of physics.  But just because you can does not mean that you should.
Setting up multiple points of lift on a manned aircraft seems like a good idea with a lot of advantages but is a Pandora's Box of problems, the not the least of which is the grave risks of single point failure.  Lift and control is totally reliant on balanced lift between multiple engines.  Lose one engine and the craft departs from controlled flight and crashes.  You better hope you have an ejection seat and/or parachute on board.
Some solutions to these problems come in the form tiltrotor designs like the Agusta Westland 609 and the V-22.  But these aircraft have multiple redundancies in their design including a common transmission system between the rotors; in case of a single engine failure, the good engine has enough power to drive both rotors.  Tiltrotors can also glide like regular airplanes and utilize control surfaces during this and cruise flight.
As a final note:  Be careful in putting too much faith in Musk and other Silicon Valley whiz kids speculating about other design endeavors.  The skills required to run a successful software company are often very different from that required to develop other artifacts like aircraft and automobiles, etc.  Musk was a genius with PayPal but nearly went bankrupt starting an electric car company and his development work with SpaceX has been punctuated with a series of spectacular failures.  He also quietly withdrew from his proposed 'Hyperloop' project, I suspect, because he really didn't understand what he was getting into there.

Thanks Carlo. Musk ad hominem aside, sounds like the fundamental issue with this approach is the fallback in case of a failure. Since electric fans are significantly smaller than regular engines, wouldn't you theoretically be able to include redundant units on board to avert this issue?

No.  Aviation gas turbines are roughly the same as the best electric motors in terms of power to weight and approx. the same size.  We have also previously covered the (at least current) impracticality of manned electric aircraft in several other posts, primarily because of the energy density of batteries compared with existing aviation fuels.  Finally my comments on Musk are based on his ideas, not on the man himself or his personal attributes.  That's not an 'ad hominem' argument.

I think your characterisation of Elon Musk is neutral to positive in my eyes. He did not start Tesla, he bought it from what he made with PayPal. By now his success has most likely bred such hubris that he now thinks the laws of physics don't apply to him. And a fawning press believes everything he says.

Sure you could, but why would you want to? There's no inherent benefit to be had in vectoring thrust as opposed to using control surfaces. It doesn't simplify things to any extent, it just makes things complicated in other ways. In order to have effective roll control you'd need to space motors out on the wings which could vector in line with the center of the aircraft, which would be complex and add weight. Also, you couldn't use thrust vectoring unless you had at least 3 engines spaced around - one in the tail for pitch and yaw, 2 in the wings for roll control.
Of course there is safety as well. A loss of power or an actuator failure, even momentarily, would probably mean a very fast, catastrophic loss of the craft and anyone inside it. Imagine a software glitch which pointed the tail engine to the left for a couple of seconds while the airplane was at cruise speed - the airplane would go side on to wind at several hundred miles per hour and most likely break up. On a VTOL approach to land one of the engines loses power, causing the craft to roll and drop the last 100 feet like a rock, destroying the craft and its occupants.

Gimbaled thrust in 'rocket speak', is equivalent to vectored thrust in 'aviation speak'.
Many modern military aircraft use thrust vectoring in order to improve performance, particularly in tight turns, the F22 being an example.
The harrier and the F35 also use thrust vectoring to enable VTOL (vertical take off and landing) and/or STOL (short take off and landing) capabilities.
Of course not only jet aircraft use thrust vectoring, nor is thrust vectoring confined only to military applications.
The Boeing V22 Osprey uses thrust vectoring, while the XTI Aircraft Company are currently developing a ducted fan VTOL civilian aircraft.
Anybody with any aviation knowledge, be it in aerodynamics or that of piloting a craft, will know that failure is unlikely to result in an uncontrolled event of any kind.
Firstly, in forward flight wings will continue to generate significant amounts of lift. Aircraft do not require engines to remain in flight, only to maintain significant airspeeds of 500mph+. Think of multiple commercial airliners that have previously experienced complete engine failure, only to glide for many miles.
Secondly, for rotary wing VTOL aircraft, a failure in any stage of flight will likely result in autorotation just as for helicopters, allowing for a great deal of control. Assuming the craft also has surfaces providing lift, this acts to provide an additional benefit.
Lastly, the very first 'control surfaces' used by the Wright brothers were warping wings.
Current control surfaces (ailerons, flaps, and the like) were originally brought into play because of the inflexibility of aircraft wings used since the early 20th century. We have used the same control surfaces ever since, but they are incredibly aerodynamically inefficient, hence why NASA has been developing alternative flexi-wings (https://www.nasa.gov/press-release/nasa-successfully-tests-shape-changing-wing-for-next-generation-aviation).
Offering a different perspective by controlling direction via engine management as oppose to manipulating airflow may improve efficiency yet further.
Beware of placing too much faith in armchair 'experts' and naysayers of online forums. Usually they have very little real experience or knowledge, let alone any vision of the future.

They do ask a couple of pertinent questions though, like how do you control the gliding aircraft without thrust. If wing morphing is the answer, why use thrust vectoring at all?

In aviation, naysaying is exactly what you need, for addressing Murphy's law. Aviation is very safe and accidents have a low probability, due to using proven methods with decades of experience. In aviation, a bug has deadly consequences.

I admit I'm not an aircraft engineer, but I have a really hard time seeing how a vectored thrust VTOL craft could autorotate.  There just wouldn't be enough airflow through those (relatively) tiny fans to produce much lift.  Helicopters can do it because they have those great big blades...

Koyovis: Search 'sailplane design' for info on how aircraft glide without thrust; aircraft do not require engines to remain in the air.
Murphy's law can be stated thus: "anything that can go wrong will go wrong". This does not leave room for naysayers, rather engineers and scientists to solve the problems of today with better solutions for tomorrow.
Jamesqf: Contrarotating fans (or multiple fans, see XTI project) will provide sufficient lift to autorotate. Contrarotating fans provide greater lifting force than equivalent non-contrarotating fans. Extra lifting surfaces will add to this.

@user23235: But doesn't it still come down to the amount of air that can pass through the area of the fan?  So taking the VTOL fans to be about the size of a typical jet engine (because if they're much larger, we have a multi-rotor helicopter), we could try imagining what would happen if such an engine, powered off but with the fan free to rotate, was dropped from a height.

jamesqf: No, I'm afraid that's completely erroneous, the reasoning being fraught with logical inexactitudes.
The amount of air able to pass through a fan duct bears no relation to the ability of the type of wing/fan within the duct to produce lift.
As stated above, a contrarotating set of fans produces greater thrust/lift than a disproportionately larger fan.
Thus a relatively small set of contrarotating fans will be able to produce as much lift as one single large fan.
See this video for a basic introduction to contrarotating propellers:
https://www.youtube.com/watch?v=Xe1g1JrRRkY

Yes I'm aware that sailplanes exist thank you. Their pilots are able to control them due to the unnecessary things like tails and rudders.

Koyovis: ? You asked "How do you control the gliding aircraft without thrust?".
A sailplane is an example of a gliding aircraft which is routinely controlled without thrust, in this specific case controlled by manipulating airflow over the wings.
What about a helicopter in autorotation, how is that controlled...? :)
What about flight using differential thrust? (here's one such example in an emergency situation: https://en.wikipedia.org/wiki/2003_Baghdad_DHL_attempted_shootdown_incident)

Koyovis: ^whoops, you edited your previous post just before I posted my response, sorry buddy :)

You won't like this answer but fixed wing VTOL aircraft with gimbals thrust actual fits many rocket/missile type of aircraft, which include:
1) High altitude anti-aircraft missiles, such as RIM-161 SM3 and S400 systems. Fins at high altitude doesn't work that well, so thrust vectoring is the main control authority at high altitude (while the body provides the lift).
2) SpaceX Falcon booster. If you argue missiles are VTO but not VL, consider the Falcon booster? VTOL, yes. Gimbals thrust, yes. Aircraft, absolutely. Fixed wing? Sort of, if you include lift body.
Within the scope of conventional manned airplanes, F35 and Harrier's reaction control is close (although differential thrust not gamble thrust).
The way you asked, "feasibility" misled many previous answers to "practicality".
For feasibility, it's a definitive yes, but for practicality, not many airplanes have done that, because of fuel economy and reliability.
Remember wings and control surfaces have a lift to drag ration greater than one, while thrust vectoring have a "lift to drag ratio" (control force to thrust lost ratio) of roughly one. For example, if your tail needs 10kn of down force, you have two options 1) use a tail plane with L/D=5 and pay for 2kn of drag, 2) divert your thrust upwards for 10kn and and lose a net thrust of 10kn.
This is the reason why thrust vectoring and differential thrust isn't used unless there are other choices, e.g. very slow(F35, Harrier, R73 AA missile) or very high (S300/400, SM3, etc), deep stall (F22, and various thrust vectoring demo and experimental airplanes, e.g. F18 HARV).


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What wing design provided the best energy retention for a fighter jet
As far as I know, tailless delta wings have the worse energy retention during a turn, is that true? If yes then why is that the case?
Also jets like F16 with root extended shallow delta wings (40% delta) are reputed to have higher energy retention, at least that's what I have concluded for the EM diagrams of F16.
Why is that the case and what wing design will be the best in energy retention?

Really, sailplane wings have the best energy retention due to high aspect ratio. But the reason you can't build a fighter jet like a sailplane is because fighter jets make sharp turns and go supersonic so you run unto structural issues due to limitations of material strength. So its not about best. It's about how much you are willing to compromise so you don't get shot down.

This is not a matter of wing type -- instead a matter of wing and aircraft design.
An aircraft designed for maneuvering might be designed such that it achieves maximum L/D in the midst of a combat turn -- say 3 g's at M=0.9 at 30,000 ft.
You of course want that L/D to be high, but beyond the wing type, this means setting the wing area such that the wing loading (W/S) is such that the above condition is met.
If an aircraft is instead designed to reach maximum L/D at 1g flight, it will likely suffer during maneuvers.  Particularly for a low aspect ratio wing that has relatively high induced drag.

So you are saying that energy retention is a matter of optimisation? I can't get how this optimisation is achieved, according to what I have found jet with same Aspect Ratio, Wing Loading, Leading Edge angle might differ significantly is performance (Turn Rate to be specific) even commercial airliners with the same engine and wing design differ in cruising efficiency, stall speed, Corner speed and all, is there any other property of a wing that I am missing?

You seem overwhelmed with the wing properties you already have -- why do you want to add more to the list?  Everything in aircraft design is a matter of optimization and tradeoffs.  You need to start with the equations and understand what they're telling you.  Look into sustained turns.

NO wing design is absolutely perfect in the configuration for what describes a fighter. In persepctive a "Delta" swept back wing makes Better interceptors and wider broader wings make better dogfighters.
THEY figured this out a long time ago. It's not a question of "Best" it's what you're thinking of in regards, Do you want, speed? or Chinking (Maneuverability)
Back then the Fighter mafia which was a group of controversial outside the Pentagon bureaucracy think group including Colonel John Boyd, Pierre Sprey...
In mid 60s they gained access to early computer mainframe and calculated a series of equations to defer mix of full embodied energy in a plane (Fuel) and equate how well that energy could be utilized either to be fast or be manueverable.
Called it EM or Energy-Maneuverability Theory. Using what essentially was early attempts "PHysics" engines in computer and man made calculation they released a 2 volume report to the US Air Force and data provided by testing of KOrean and Vietnam era aircraft provided data Air Force to design criteria for it's next generation of planes which became the "Teen Series" which gave rise to the F15 and F16.

What is "chinking"?

This could use an edit for clarity, reduction of raNdom caPs and readability

@quietflyer "Jinking" is the more usual spelling.  "To move quickly or unexpectedly with sudden turns and shifts" (Merriam-Webster).


text:
How/why does engine torque and horsepower affect aircraft speed?
I'm curious as to why torque and horsepower affect an aircraft's speed. AFAIK, it's only the engines RPM that affect speed, right?
Does this mean an engine's torque/horsepower only affects the speed it can effectively operate at or am I wrong?

Horsepower is what propels an airplane through the air; more horsepower means more speed. Power is neither torque nor RPM by itself: it is their product. (Torque)x(rpm)x(conversion factor) = horsepower.

how does horsepower propel the airplane through the air faster?

@alekslacroix, horsepower is a measure of how fast a machine can perform work. in a flying airplane, the work is used to overcome air friction, which is created by moving through the air.

so if the horsepower of an engine was too low the propellers would slow down due to air friction?

@alekslacroix You are correct. This tiny motor and this F1 engine are both rated for 15,000RPM - and yet one outperforms the other. Clearly, RPM isn't the only factor affecting performance; torque is equally important. You cannot replace the V6 with the tiny motor and expect the same performance from it - all that RPM is of no use if it doesn't have enough torque to turn the wheels.

Similarly, to turn a propeller through the air, you need torque - thousands of foot-pounds of torque. Since both RPM and torque determine the performance of an engine, we use the product of the two, known as horsepower, to compare the performance of engines.

Horsepower and torque affects RPM.
For a given torque, an aircraft needs greater RPM to achieve greater speed. And that means more horsepower.
For a given RPM, an aircraft with a higher blade pitch travels faster, but this requires more torque, due to the more sideways direction of the force in the blades. More torque means more horsepower.

Perhaps, in context, this example may be helpful.  We have, say, a piston engine turning a propeller sufficiently to produce enough thrust from that propeller to maintain an aircraft in flight.  Consequently, the two things we have are a) torque horsepower from the piston engine, and b) thrust horsepower from the propeller.  The propeller is not perfectly efficient, but nearly so.  Therefore, we can state with confidence the following relationship
$$Efficiency = \frac{Thrust Horsepower}{Torque Horsepower}$$  When we fly the plane, we set the throttle to a specific fuel flow at a specific manifold pressure to maintain a specific rpm.  The propeller being turned by that engine does its thing and produces the thrust required for flight.  So how much thrust is that, you may ask.  Well, the thrust is equal to the drag produced by the aircraft.  We can measure the drag or specifically estimate the drag for a given aircraft velocity, and thereby, determine the power produced by that propeller.  The relationship is simply $$\frac{D \cdot\ V}{550} = Thrust Horspower$$ where D is the drag in pounds and V is the velocity in feet per second.  If our propeller is 89 percent efficient, then the torque horspower required to develop that thrust would be 1/0.89 times the thrust horse power, or alternatively, since we know the power setting of the piston engine turning that propeller, and we know the propeller efficiency, we can estimate very closely the thrust horsepower as 89 percent of the torque horesepower.  Also notice that this is all balanced on the velocity of the aircraft.  Hence, the following must be true -
$$V = \frac{Thrust Horsepower \cdot\ 550}{D}$$ or, specifically,
$$V = \frac{Torque Horsepower \cdot\ 550 \cdot\ 0.89}{D} .$$

Horsepower is a measure of the total power output, this is proportional to RPM multiplied by torque.  How this relates to static thrust depends on the diameter and pitch of the propeller; a bigger diameter will move more air and a steep pitch will move that air faster.  An aircraft designer will try to optimise these so that the full power of the engine can be used at an appropriate engine speed.  This is complicated by many factors.  The combination of pitch and RPM gives the speed at which the propeller will try to move the air, as the aircraft nears this speed the thrust will decrease and so a faster aircraft will call for a steeper pitch.

For the same reason a car's engine and its transmission affect its speed. You can have an engine in a car producing constant horsepower that never changes, but given a flat road, its speed will depend on what gear has been selected in the transmission. More or less horsepower doesn't always translate to more or less speed.


text:
How do the landing gear support so much mass?
The first thing that one notices after seeing an aircraft, is that it's quite a heavy flying machine. And so, one would expect monumental tires on these vehicles, yet they don't exist.
How is it that such small tires cannot only carry the weight of planes like 747s and A380s, but also prevent getting too damaged during touchdown?
Is there a certain mechanic that prevents the collapse of landing gear and bursting of the tires?

The huge energy absorbed when breaking gives an idea of what the wheels must deal with, see the excellent answers to How are brakes cooled on heavy aircraft?

I think your question is based on a perception that isn't shared by all.
Yes, airplanes can be very large, and numerically quite heavy when loaded at max gross, but appearances can be deceiving and they are actually fairly light for their size.  They are mostly hollow, and designed to be as light as possible... because they need to fly after all!  Consider that they are made of thin (compared to overall size) aluminum, and that much of the overall size you perceive is in the volume of the fuselage, which contains a lot of air.
Realize too that the visual appearance of size doesn't always match well with actual density or mass.  Mentally scaling up other more familiar wheeled vehicles or die cast airplane models may create a subconscious illusion that airplanes are much more massive than they actually are.
I don't have the engineering chops to math it out for you, (if you are looking for strength calculations) I can only say that my perception has never been that the tires are too small.

Is there a certain mechanic that prevents the collapse of landing gear and bursting of the tires?

Yes, a team of design engineers will spend years making sure the gear is strong enough.

Despite the fact that LCCs pack as many people as possible into the cabin, even a full passenger aircraft is probably more air than "stuff" at take-off.

In addition to the number of tires and strength of the landing gear, an important factor is pressure. Typical car tire pressure is on the order of 35 PSI. Typical large airplane tire pressure is on the order of 200 PSI!
While I don't know all the engineering behind it, my understanding is that higher pressure = support more weight. Consider that a small "donut spare" will be inflated to 60 PSI rather than the 35 PSI of a typical car tire, because that allows the smaller spare tire to, within certain constraints, handle as much load as a regular tire.

Yeah: it is the pressure of the gases which support the aircraft, so you need enough pressure to support the weight. OTOH with too much pressure we have something like solid tires which do not damp irregularities of runways/taxiways (and landing).

"my understanding is that higher pressure = support more weight." Indeed, pressure opposes weight. Weight is distributed on a small contact area with ground, pressure (in Pa, 1 psi = 6894 Pa) times this area (in m²) must be equal to weight (in kg) acting on the wheel. Thus 10 psi can counter about 6 tons on an area of 30x30cm.

Most civil airways require a pressure below 120psi otherwise the tarmac might get damaged. 200psi is more common for an aircraft carrier. "While I don't know all the engineering behind it" that's very easy: force is pressure times contact surface; contact surface is mainly dependent on the physical dimensions of the tire $\Rightarrow$ higher pressure, smaller tire.

Consider also, that heavy road-going truck tires (semi tractor-trailers, dump trucks, etc.) have tire pressures in the 100-120 PSI range to support weights upwards of 80,000 lbs.

That's quite easy: you split the weight among several wheels (22 for an A380 for example) and design the landing gear strong enough (their weight can be as high as 8% of the empty weight, corresponding to around 20t for an A380).
This nice presentation (pdf) gives an overview of the landing gear development for the A380.

While it seems nearly impossible for the landing gears to be able to handle the aircraft load, there is a lot more that stays hidden. 3 of them are:

Hydraulics that dampen a lot of load (dynamic mostly for the Max Landing Weight), and,
The kinematics, that distribute the load across multiple linkages
The last one hiding in plain sight is the number of wheels x number of landing gears. While a small/medium sized aircraft may have 2 wheel long landing gears, a Boeing 777 could have an additional wheel or a larger aircraft like the A380 could have multiple of them (even on the wings!!)

Always remember manufacturers need to comply to stringent standards like AC23.1309 and at times also develop the products based on recommended practices like ARP4754 to ensure the design is absolutely safe (almost) with a very high factor of safety (we are talking about a number greater than 1.2)


text:
What type of aircraft is this?
Flew very low in circles around my house this morning. I live near Fox Harbour airport in Wallace, Nova Scotia
Click to embiggen

Wow. Thank you for all of that information. It was really something to stand under it and have it pass over head so many times.

I love this site (and many of the sister sites, among them Science Fiction and Fantasy for story identification!) because there is always somebody there who can answer these nagging questions (I had a similar one, and just 20 years ago we would never have known). Kudos to Bianfable here, and have a nice summer in Nova Scotia.

That is a de Havilland Canada Dash 7, a 4-engine turboprop STOL aircraft.
You can see the flight this morning using the Flightradar24 Playback feature:

Flightradar24 does not show the registration of this aircraft, but ADS-B Exchange shows it as C-GCFR:

(jetphotos.com)
This aircraft is operated as a surveillance aircraft by Transport Canada (note the special equipment on top of the fuselage in the picture above):

The iconic red planes of Canada's National Aerial Surveillance Program (NASP) are a pillar in helping Canada stay safe. Since 1991, we've grown — adding new aircraft, new technology and new people — all with the goal of preventing pollution in our waters, protecting our marine environment and endangered marine life, and ensuring a safe and efficient transportation industry along our massive and varied coastlines. [...]
Today's NASP fleet includes aircraft owned and operated by Transport Canada based strategically across Canada for effective monitoring and rapid response to incidents:

Dash 8 based in Moncton serving Central and Eastern Canada [...]

NASP operations rely on a suite of remote sensing equipment that includes:

Side Looking Airborne Radars
Infrared / Ultraviolet Line Scanners
Electro-Optical Infrared Camera Systems
Automatic Identification Systems
Satellite Communications Systems
Geo-coded Digital Camera Systems

(Transport Canada)

Good find, also interesting to see the special configuration with the 'bubble' on top. Link to NASP activities page

@ROIMaison Good point, I added some info on the equipment used from Transport Canada's website.

Only 113 Dash 7s were built. It was a bit of a marketing flop, designed to exploit a prediction that MLS would allow operations from "stub runways" at super busy airports, and which went nowhere. Too much focus on STOL, not enough on efficiency, esp with 4 PT6-50s, chosen because were weren't any up-to-date TPs with the required power to make a Twin work. The Dash 8 was basically a do-over to provide what the regional market really wanted, made possible by P&W's development of the PW-120.


text:
What is the correct cruise operating temperature for the PT6A-60 in a King Air 350?
Why are King Air 350 PT6 operating temps the way they are? I’ve been taught to cruse at 785 as per Beech/Textron maintenance recommendations. My non typed not really qualified FO claims I am ruining Engine life by running these temps. He likes to cruse in the 750 range. I don’t really care so long as he doesn’t kill me.
But to say I’m destroying the engines by running at 785 is a little bit annoying.
Am I actually damaging the aircraft by operating at 785?

First off, why isn’t your F/O qualified or typed? Secondly, why are you listening to him if he’s not qualified in the aircraft?

"Ruining" is a bit of a stretch if you are following Beech/Textron recommendations.  Show him that and end the conversations.  However, running cooler will ALWAY result in longer life, so his argument is not without some merit...

Extraordinary claims require extraordinary evidence. Can this guy cite information, either from Beechcraft or Pratt & Whitney Aircraft of Canada confirming this? If he's not typed in the thing, I'm not buying it otherwise.

Actually, you are both wrong. If you look at the BE300 cruise power tables you will see that you are supposed to set torque for power (based on P.Alt., temp. and weight) and not ITT.
When you purchase a new King Air 300/350 it comes with a Pratt and Whitney 'how to care for your engine' booklet which explains why this is so.
Using ITT to set power can result in premature replacement of the power turbine blades (perhaps even before the mid time hot section check is due) and/or the engine not developing its full rated power towards the end its life (which renders the performance section of the handbook useless).
The engine is designed to use constant torque settings throughout its life. The resulting ITT's will slowly increase as the engine ages, but will still remain within operating limits. This will result in the engine always producing its full rated power output (which is what the performance section of the handbook is based upon).
Setting ITT for power is simply a guessing game. If you set them low, then the engine is probably just fine, but you have no clue as to whether you are meeting the power requirements necessary to make the performance tables valid... which is not fine. One day (with perhaps 3,000 hours on the engines) you calculate your single engine climb out of Eagle in IMC as 500 fpm, an engine quits, you begin the single engine alternate departure procedure (which may require 400 fpm) and you only manage 300 fpm! and then it's "hello granite".